Calculate the change in pH of water when `0.01` mole of NaOH are added in 10 litre water.

Explanation

To calculate the change in pH of water when 0.01 moles of NaOH are added to 10 liters of water, we can follow these steps:

### Step 1: Calculate the concentration of NaOH
First, we need to find the concentration of NaOH in the solution after it is added to the water.

Given:
- Moles of NaOH = 0.01 moles
- Volume of water = 10 liters

Concentration (C) of NaOH is given by the formula:
\[ C = \frac{\text{moles of solute}}{\text{volume of solution in liters}} \]

So,
\[ C = \frac{0.01 \text{ moles}}{10 \text{ liters}} = 0.001 \text{ M} \]

### Step 2: Determine the pOH of the solution
Since NaOH is a strong base, it completely dissociates in water:
\[ \text{NaOH} \rightarrow \text{Na}^+ + \text{OH}^- \]

Thus, the concentration of hydroxide ions \([OH^-]\) will also be 0.001 M.

The pOH is calculated using the formula:
\[ \text{pOH} = -\log[OH^-] \]

Substituting the value:
\[ \text{pOH} = -\log(0.001) = 3 \]

### Step 3: Calculate the pH of the solution
We know that:
\[ \text{pH} + \text{pOH} = 14 \]

So, we can find the pH:
\[ \text{pH} = 14 - \text{pOH} = 14 - 3 = 11 \]

### Step 4: Determine the change in pH
The initial pH of pure water is 7.

Now, we can find the change in pH:
\[ \text{Change in pH} = \text{Final pH} - \text{Initial pH} = 11 - 7 = 4 \]

### Final Answer
The change in pH when 0.01 moles of NaOH are added to 10 liters of water is **4**.

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